Thermal Physics Solutions

Thermal Physics Solutions 1. PV  NkT PV kT

 N

 2.0 10    5.0 10    1.38 10    300  4

5

23

 2.4  10 22

2(i) T 

 PV  T  273.16  PV  tr



PT  273.16 Ptr



4870  273.16 2680

since V is constant

=496 K (ii) Real gases behave like ideal gases at low pressure and the ideal gas equation is exactly what the thermodynamic scale is based on 3. PV  nRT V



nRT P 1  8.31  300 1.000  105

 0.025 m3 mol -1



 

Volume of 1 mol of molecules = 2.000  10 29  6.02  10 23



 1.204  105 m3

Fraction of gas that is empty =



0.025  1.204  105



0.025  0.9995

4. Initial

Final

A PA1 = Patm nA1

B PB1 = Patm nB1

TA1 = 100˚C = 373.15 K

TB1 = 20˚C = 293.15 K

A PA2 = P nA2

B PB2 = P nB2

TA1 = 20˚C = 293.15 K

TB1 = 20˚C = 293.15 K

Compare initial and final states of A PA1VA1 PA2 VA2  n A1TA1 n A2 TA2 

Patm P  n A1 (373.15) n A2 (293.15)

 P

using PA1 = Patm and VA1  VA2

n A2 (293.15)  Patm ---------------- (1) n A1 (373.15)

*(Note: from equation (1), we find that we need an equation to express nA1 in terms of nA2 only in order to solve for P)

For initial state, PA1VA1 PB1VB1  n A1TA1 n B1TB1  n A1TA1  n B1TB1  n A1 (373.15)  n B1 (293.15) since PA1 = PB1 and

-------- (2)

VA1  VB1

For final state, PA2 VA2 PB2 VB2  n A2 TA2 n B2 TB2  n A2  n B2 -------- (3) since PA2 = PB2 = P , VA2  VB2 and

TA2  TB2

The total number of moles of molecules in the initial state is the same as the total number of moles of molecules in the final state: n A1  n B1  n A2  n B2

----------------- (4)

* Note that we can obtain an equation involving nA1 and nA2 only if we use (2),(3) and (4) Sub (3) into (4) n A1  n B1  2n A2 ----------------- (5)

Sub (2) into (5) n A1 

n A1 (373.15)  2n A2 293.15

 n A1 

2n A2 373.15    1  293.15   

 n A1  0.8799 n A2

------------------ (6)

Sub (6) into (1) P

n A2 (293.15)  Patm n A1 (373.15)

P

n A2 (293.15)  (1.013  105 ) 0.8799n A2 (373.15)

= 9.04  104 Pa

5. Force Area F = A mg = A mgh = Ah  m  = hg  Ah  m = hg V = ρhg

Pressure =

(i) (ii)

At constant pressure P, V1 V2  T1 T2 

0.00783 0.0308  273 T1

 T1  69.4K

6(i)

A

From kinetic theory of matter, matter is made up of particles that are in continuous random motion. Pressure is brought about due to the collisions of the molecules with the walls of the container it is in. Not in syllabus

c(i)

(iii)

mg

273 K = -0.15˚C 69.4 K = -203.7˚C

1 m v 2

2







1 2 3.6  1025  4.6  1026   270  2

= 60400 J (3.s.f.)

(ii)





2 1 1 2 m  v  240   3.6  1025  4.6  1026   270  240  2 2

= 215 kJ (3.s.f.)

(iii) Internal energy of the gas = kinetic energy of the gas = 215 kJ

(since gas is ideal)

7 (a) PV  nRT n



 

5 4 PV 1.00  10  5  10  RT 8.31  300



= 0.02 mol

(b)

Q = mc∆θ Q = ncn∆θ

(m = mass, c = specific heat capacity or heat capacity per unit mass) (n = no of moles, cn = molar heat capacity or heat capacity for 1 mol of substance)

Q = 0.02 21 (1500 – 630) = 365.4 kJ (c) Since there is no change in volume, there is no work done by the gas. (d) Find pressure at C first PB PC  TB TC

 1.50 10    6

630

PC 1500

 1.50 10   1500  6

 PC

630

 PC  3.57  106 Pa

Pressure at C is 3.57  106 Pa

Use PD VD PC VC  TD TC  PD 

TD PC VC  VD TC

 PD 

680 (3.57  106 )  (7  105 )  1500 5.00  104

 227 kPa

Pressure at D is 227 kPa (e)

P105/ Pa C (1500 K)

35.7

15.0

(630 K) B

2.27

D (680 K)

1.00

A (300 K) 7.00  10

-5

5.00  10

-4

V/m3